Henry Ernest Dudeney/Modern Puzzles/64 - Dividing by Eleven/Solution

Modern Puzzles by Henry Ernest Dudeney: $64$

Dividing by Eleven

If the $9$ digits are written at haphazard in any order,

for example $4 \ 1 \ 2 \ 5 \ 3 \ 9 \ 7 \ 6 \ 8$, what are the chances that the number that happens to be produced will be divisible by $11$ without remainder?

The number I have written at random is not, I see, so divisible, but if I had happened to make the $1$ and the $8$ change places it would be.

Solution

$115$ to $11$ against a haphazard arrangement being divisible by $11$.

Proof

Let $\sqbrk {abcdefghi}$ be a haphazard arrangement of the $9$ digits.

We set:

$m = a + c + e + g + i$

$n = b + d + f + h$

From Divisibility by 11, $\sqbrk {abcdefghi}$ is divisible by $11$ if and only if:

$m - n = 11k$

for some integer $k$.

We have:

\(\ds \)

\(\)

\(\ds 25\)

\(\ds \)

\(=\)

\(\ds \paren {5 + 6 + 7 + 8 + 9} - \paren {1 + 2 + 3 + 4}\)

\(\ds \)

\(\ge\)

\(\ds m - n\)

\(\ds \)

\(\ge\)

\(\ds \paren {1 + 2 + 3 + 4 + 5} - \paren {6 + 7 + 8 + 9}\)

\(\ds \)

\(=\)

\(\ds -15\)

Moreover, since $m + n = 45$ is odd, so is $m - n = \paren {m + n} - 2n$.

Hence we can only have $m - n = \pm 11$, which give the following cases when solved:

Case $1$: $m = 28, n = 17$

Case $2$: $m = 17, n = 28$

For Case $1$, the following $9$ quadruples of distinct digits give $n = 17$:

$1, 2, 5, 9$

$1, 2, 6, 8$

$1, 3, 4, 9$

$1, 3, 5, 8$

$1, 3, 6, 7$

$1, 4, 5, 7$

$2, 3, 4, 8$

$2, 3, 5, 7$

$2, 4, 5, 6$

For Case $2$, the following $2$ quintuples of distinct digits give $m = 17$:

$1, 2, 3, 4, 7$

$1, 2, 3, 5, 6$

Hence there are exactly:

$\paren {9 + 2} \times 4! \times 5! = 31680$

numbers, out of $9! = 362880$ possibilities that our number is divisible by $11$.

This gives a probability of:

$\dfrac {31680} {362880} = \dfrac {11} {126}$

which is exactly $115$ to $11$.

$\blacksquare$

Sources

• 1926: Henry Ernest Dudeney: Modern Puzzles ... (previous) ... (next): Solutions: $64$. -- Dividing by Eleven
• 1968: Henry Ernest Dudeney: 536 Puzzles & Curious Problems ... (previous) ... (next): Answers: Digital Puzzles: $115$. Dividing by Eleven